Number of occurarece of each char in a string

Question

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Count occurrences of each unique character

How can I get the number of occurrence of each char in a string? for example:

"stackoverflow": s:1 t:1 a:1 c:1 k:1 o:2 v:1 e:1 r:1 f:1 l:1 o:1 w:1

and what have you googled? ;) – Peter Lawrey Jan 06 '12 at 16:21 — Peter Lawrey, Jan 06 '12 at 16:21

score 3 · Answer 1 · answered Jan 06 '12 at 16:35

Guava to the rescue! I can't believe it, this is a java one-liner!

Multiset<Character> distintCharsAndCount = HashMultiset.create(
    Chars.asList("ssssssssstackoverflow".toCharArray())
);

Chars.asList helps us convert the string to a real collection of non-primitives. MultiSets are the perfect structure for what you're after: it's a Set that keep distinct elements, but also keeps the number of occurances in the set.

Access it this way:

int sCount = distintCharsAndCount.count('s'); //sets it to 9

score 2 · Answer 2 · answered Jan 06 '12 at 16:21

Create HashMap with keys as a Character and Integer count as value.

HashMap<Character, Integer> hm = new HashMap<Character, Integer>()
for (int i = 0; i < str.length; i++) {
    if (hm.get(str.charAt(i))) {
        int temp = hm.get(str.charAt(i));
        hm.put(str.charAt(i), ++temp);
    } else {
        hm.put(str.charAt(i), 1);
    }
}

score 1 · Answer 3 · answered Jan 06 '12 at 16:20

1

I would suggest using a map from character to count and iterating over the character array adding to the map or the count as you go along

answered Jan 06 '12 at 16:20

James

264
1
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13

score 1 · Answer 4 · answered Jan 06 '12 at 16:24

1

Map<Character, Integer> chars = new HashMap<Character, Integer>();
for (int i = 0; i < str.length; i++) {
    char c = str.charAt(i);

    Integer count = chars.get(c);
    if (count == null) {
        chars.put(c, 1);
    } else {
        chars.put(c, count + 1);
    }
}

answered Jan 06 '12 at 16:24

Jon Newmuis

25,722
2
45
57

As this is homework and the OP has not shown any effort would it not have been better to point him/her in the right direction as opposed to doing it for them? – Darren Burgess Jan 06 '12 at 16:36

score 1 · Answer 5 · answered Jan 06 '12 at 16:26

A simple but intuitive way that I have done this in the past is to cast the chars as ints and then just use an array. Using a Map is a better approach, but this sounds like a homework problem and using casting and an array is a bit more logical (imo) for a beginner.

int counts[] = new int[26];
String s = 'stackoverflow';
s = s.toLowerCase();
for (int i = 0; i < s.length(); i++) {
    int val = (int)(s.charAt(i)) - 97;
    counts[val]++;
}

score 0 · Answer 6 · answered Jan 06 '12 at 16:22

0

You could parse the string and create a Map<Character, Integer> so that the integer (value in the map) represents how many times the character c (the key in the map) appears in the string.

answered Jan 06 '12 at 16:22

Wilmer

1,025
5
9

score 0 · Answer 7 · answered Jan 06 '12 at 16:24

public static Map<Character, Integer> count(String s) {
  Map<Character, Integer> result = new HashMap<Character,Integer>();
  for (int i = 0; i < s.length(); ++i) {
    char c = s.charAt(i);
    Integer n = result.get(c);
    result.put(c, n == null ? 1 : n + 1);
  }

  return result;
}

score 0 · Answer 8 · answered Jan 06 '12 at 16:34

String input = "stackoverflow";

// desired output: s:1 t:1 a:1 c:1 k:1 o:2 v:1 e:1 r:1 f:1 l:1 o:1 w:1
Map<Character,Integer> counts = new HashMap<Character, Integer>
( Math.min(input.length(),Character.MAX_VALUE) / 2 );
char ch;
for (int i=0; i<input.length(); ++i)
{
  ch = input.charAt(i);
  if ( !counts.containsKey(ch)) { counts.put(ch,0); }
  counts.put(ch, counts.get(ch)+1 );
}

for (Map.Entry<Character,Integer> entry : counts.entrySet() )
{
  out.print( entry.getKey() );
  out.print( ":" );
  out.print( entry.getValue());
  out.print( " " );
}

Number of occurarece of each char in a string

8 Answers8